My Math Forum order of the product of two disjoint cycles is the least common multiple

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 January 19th, 2017, 01:51 PM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 order of the product of two disjoint cycles is the least common multiple Hi, I have this exercise, but, I have some problems to develop it. Prove that if $\alpha$ represents an r-cycle in $S_n$ and $\beta$ represents an s-cycle, and $\alpha$ and $\beta$ are disjoint, then $o(\alpha \beta) = [r,s]$ (i.e. the least common multiple of $r$ and $s$). END I don't know how to resolve it. So let's see what we have: we have a group $S_n$, and its elements are permutations called as $\alpha, \beta, ...$, permutations can be written in form of cycle, and we take two cycle from the group. $\alpha = (a_1 a_2 ... a_n), \beta = (b_1 b_2 ... b_n)$ by definition two cycles $(a_1 a_2 ... a_n), (b_1 b_2 ... b_n)$ are disjoint if $a_i \ne b_j$ with $i = 1, ..., r; j = 1, ... , s$. For example $(1 2 4), (3 5 6)$ are disjoint, but, $(1 2 4), (3 4 6)$ are not. Also disjoint cycles commute. by definition, the order of $\alpha \beta$ would be the integer $n$ as: $(\alpha \beta)^n = (1) = e$ If I understand, $r,s$, are respectively the number of elements within the cycles $\alpha, \beta$, so it can be possible that $r \ne s$. So, in what way can I prove that it is the least common multiple of r ans s? please, can you give me any help? thanks!
 January 19th, 2017, 06:55 PM #2 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 Start by proving that if two permutations are disjoint, then, when raised to a power, the resulting permutations are still disjoint. Then argue that a product of two disjoint permutations can't equal 1 unless they're both equal to 1. Next suppose $(\alpha \beta)^t = 1$. What can you say about $\alpha^t$ and $\beta^t$? Deduce that $r|t$ and $s|t$. Verify that $(\alpha \beta)^{[r,s]} = 1$. Conclude that $o(\alpha \beta)=[r,s]$. Thanks from beesee

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### prove the order of 2 disjoint cycles is the lcm

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